Question: A particle moves along the curve $y=2x^2+2$ so that that the $x$ -coordinate is increasing at a constant rate of $4$ units per second. What is the magnitude (in units per second) of the particle's velocity vector when the particle is at the point $(0,2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $16$ (Choice B) B $4$ (Choice C) C $2$ (Choice D) D $34$
Explanation: Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=4$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(0,2)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(0,2)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=16x$ Finding $\dfrac{dy}{dt}$ at $(0,2)$ The expression for $\dfrac{dy}{dt}$ only depends on the particle's $x$ -coordinate, which in our case is ${x}={0}$ : $\begin{aligned} \dfrac{dy}{dt}&=16({0}) \\\\ &=0 \end{aligned}$ Therefore, the particle's velocity vector at the point $(0,2)$ is $(4,0)$. Finding $||(4,0)||$ $|(4,0)||=4$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(0,2)$ is $4$ units per second.